How much CaF2 (in grams) is formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate?

Question

How much CaF2 (in grams) is formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate? Please help me

Answer 1

The net ionic equation for this reaction is 2F^-(aq) + Ca²⁺(aq) ==> CaF₂(s)

moles F^- present = 0.0157 L x 0.612 mol/L = 0.009608 moles F^-

moles CaF₂ formed = 0.009608 moles F^- x 1 mol CaF₂/2 mol F^- = 0.004804 moles CaF₂

mass CaF₂ = 0.004804 moles x 78.07 g/mole = 0.375 g CaF₂ formed